#### Question

If p, q are real and p ≠ q, then show that the roots of the equation (p − q) x^{2} + 5(p + q) x− 2(p − q) = 0 are real and unequal.

#### Solution

The quadric equation is (p − q) x^{2} + 5(p + q) x− 2(p − q) = 0

Here,

a = (p - q), b = 5(p + q) and c = -2(p - q)

As we know that D = b^{2} - 4ac

Putting the value of a = (p - q), b = 5(p + q) and c = -2(p - q)

D = {5(p + q)}^{2} - 4 x (p - q) x (-2(p - q))

= 25(p^{2} + 2pq + q^{2}) + 8(p^{2} - 2pq + q^{2})

= 25p^{2} + 50pq + 25q^{2} + 8p^{2} - 16pq + 8q^{2}

= 33p^{2} + 34pq + 33q^{2}

Since*, P* and *q* are real and p ≠ q, therefore, the value of D ≥ 0.

Thus, the roots of the given equation are real and unequal.

Hence, proved

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#### APPEARS IN

Solution If P, Q Are Real And P ≠ Q, Then Show that the Roots of the Equation (P − Q) X2 + 5(P + Q) X− 2(P − Q) = 0 Are Real and Unequal. Concept: Nature of Roots.